Changing the order of integration in $\int_0^1 \int_x^1 f(x, y) dy dx yields$:
Region: 0 <= x <= 1 and x <= y <= 1. Region is a triangle bounded by y=x, x=0, y=1. Reversing order: 0 <= y <= 1 and 0 <= x <= y.
Compute the area of the region bounded by $y = x^2$ and y = 4x.
Intersections: $x^2$ = 4x => x = 0 and x = 4. Area = $\int_0^4 (4x - x^2) dx = [2x^2 - x^3/3]_0^4 = 32 - 64/3 = 32/3 = 10.666$... $\approx 10.67$.
The Jacobian J = $\frac{\partial(x, y)}{\partial(r, \theta)}$ for polar coordinates x = r $\cos \theta, y = r \sin \theta$ is:
J = det([[cos theta, -r sin theta], [sin theta, r cos theta]]) = r $cos^2$ theta + r $sin^2$ theta = r.
Evaluate the double integral $\iint_R dx dy$ where R is the circle $x^2 + y^2$ $\le 9$.
The double integral equals the area of a circle of radius R=3. Area = $\pi R^2 = 9\pi \approx 9 * 3.14159 = 28.27$.
Compute $\int_0^{\pi} \sin^2(x) dx$.
Using identity $sin^2$ x = (1 - cos 2x)/2. Integral = [x/2 - $\sin(2x)/4]_0^{\pi} = \pi / 2 \approx 3.14159 / 2 = 1.5708 \approx 1.57$.
